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CIE A-Level Maths Study Notes

1.6.1 Binomial Expansion

The binomial expansion is a fundamental concept in Maths, offering a systematic approach to expanding expressions raised to a power. This method is particularly valuable for simplifying expressions with high exponents, where direct multiplication is not feasible.

Understanding the Binomial Theorem for Positive Integer Exponents

The binomial theorem facilitates the expansion of expressions in the form (a+b)n(a+b)^n, where nn is a positive integer. The expansion involves a sum of terms featuring binomial coefficients.

Binomial Coefficients and Factorial Notation

  • Binomial coefficients are represented as (n r)\left(\begin{array}{c}n \ r\end{array}\right), where nn is the exponent and rr ranges from 0 to nn.
  • These coefficients, also written as nCr^nC_r, are calculated using the formula n!r!(nr)!\frac{n!}{r!(n-r)!}, where n!n! denotes the factorial of nn.

Expansion Formula

The binomial expansion of (a+b)n(a+b)^n is: (a+b)n=k=0n(n k)ankbk(a+b)^n = \sum_{k=0}^n \left(\begin{array}{c}n \ k\end{array}\right) a^{n-k} b^k This summation notation provides an alternative way to express the expanded form.

Key Patterns in Binomial Expansion

  1. Symmetry: nCr=nCnr^nCr = ^nC{n-r}.
  2. Boundary Values: nC0=nCn=1^nC_0 = ^nC_n = 1 and nC1=nCn1=n^nC1 = ^nC{n-1} = n.
  3. Pascal's Triangle is a useful tool for visualising these patterns, withnnindicating the row (starting from 0) and rr the position from the left (also starting from 0).

Worked Examples

Example 1: Basic Expansion

Expand the expression (a+b)5(a+b)^5.

Solution:

Applying the binomial theorem:

(a+b)5=a5+(5 1)a4b+(5 2)a3b2+(5 3)a2b3+(5 4)ab4+b5(a+b)^5 = a^5 + \left(\begin{array}{c}5 \ 1\end{array}\right) a^4b + \left(\begin{array}{c}5 \ 2\end{array}\right) a^3b^2 + \left(\begin{array}{c}5 \ 3\end{array}\right) a^2b^3 + \left(\begin{array}{c}5 \ 4\end{array}\right) ab^4 + b^5=a5+5a4b+10a3b2+10a2b3+5ab4+b5= a^5 + 5a^4b + 10a^3b^2 + 10a^2b^3 + 5ab^4 + b^5

Example 2: Expansion with Coefficients

Expand the expression (3x+2)4(3x+2)^4.

Solution:

Using the binomial theorem:

(3x+2)4=(3x)4+(4 1)(3x)32+(4 2)(3x)222+(4 3)(3x)23+24(3x+2)^4 = (3x)^4 + \left(\begin{array}{c}4 \ 1\end{array}\right)(3x)^3 \cdot 2 + \left(\begin{array}{c}4 \ 2\end{array}\right)(3x)^2 \cdot 2^2 + \left(\begin{array}{c}4 \ 3\end{array}\right)(3x) \cdot 2^3 + 2^4=81x4+427x32+69x24+43x8+16= 81x^4 + 4 \cdot 27x^3 \cdot 2 + 6 \cdot 9x^2 \cdot 4 + 4 \cdot 3x \cdot 8 + 16=81x4+216x3+216x2+96x+16= 81x^4 + 216x^3 + 216x^2 + 96x + 16

Example 3: Finding Specific Coefficients

Find the coefficient of x3x^3 in the expansion of (1+2x)8(1+2x)^8.

Solution:

Identifying the term with x3x^3 in the binomial expansion:

(1+2x)8=+(8 3)183(2x)3+(1+2x)^8 = \ldots + \left(\begin{array}{c}8 \ 3\end{array}\right) 1^{8-3} \cdot (2x)^3 + \ldots =+561(2x)3+= \ldots + 56 \cdot 1 \cdot (2x)^3 + \ldots =+568x3+= \ldots + 56 \cdot 8x^3 + \ldots

Therefore, the coefficient of x3x^3 is 448.

Dr Rahil Sachak-Patwa avatar
Written by: Dr Rahil Sachak-Patwa
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Oxford University - PhD Mathematics

Rahil spent ten years working as private tutor, teaching students for GCSEs, A-Levels, and university admissions. During his PhD he published papers on modelling infectious disease epidemics and was a tutor to undergraduate and masters students for mathematics courses.

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